Question

Find the vector and scalar equations for the line in \[\mathbb{R}^3\] that contains the Point P(-1,6,0) and is orthogonal to the plane 4x - z = 5

Can anyone help me with this question?

I understand that to tell if two vectors are orthogonal you need to take the dot product and determine if they are equal to zero

Where do I start with this problem can anyone point me to some resources that will help me to solve it, please do not give me the answer.

Answer 1

If you have a plane given in normal form, like in that problem, you can immediately tell it's normal vector component, since the line in R^3 is supposed to be orthogonal to it, means that the normal vector of the plane is a direction vector of the line itself.

Answer 2

can I assume y = 0, so the normal vector for 4x - z = 5, would be
n = [4 0 -1]^T

Answer 3

exactly, that's what I would have done.

Answer 4

where do I go from there though?

Answer 5

The normal is orthogonal right

Answer 6

To make a line unique you need to match it to the given point.
\[r_x=0P + t v\]
where v=n so you just need to plugin your point into that equation.

Answer 7

The normal is orthogonal to the plane itself, and since the line has to be orthogonal to the plane that means that the normal vector of the plane is the direction vector of the line. The direction vector is usually described as v and t is a linear scalar quantity.

Answer 8

what is rx and 0P in this case though?

Answer 9

sorry if I'm a bit slow ha

Answer 10

Hmm pardon me, I am from Europe and we have slight different notations for things here in case you're from the USA.

0P indicates a point, a point that is on the line, r_x only means (x,y,z) in three dimensions or (x,y,z,....,n) in n-dimensions.

So your equations will look like that for the x component:

x=-1+4t

Answer 11

can you do the same for the y= and z= components ?